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u(x) = 0, and therefore is ju-negligible (13.14.1), hence v-negligible by hypothe-
sis. On the other hand, v" = vq>At hence ((13.14.1) and (13.6.3)) the relation
v(A) = 0 is equivalent to v" being A-negligible, and therefore implies that
v" . A = 0 (13.14.4). To show that (d) implies (a), it is enough to remark that
the condition (d) signifies that v"  A = 0, and hence (13.14.4) the set A is
1-negligible. Hence if we put g(x) = v(x)/u(x) at points where u(x) > 0, and
g(x) = 0 elsewhere, we shall have v(x)  g(x)u(x) at all x $ A, hence almost
everywhere with respect to L Consequently (13.14.5), g is locally /z-integrable,
and we have v = g  /x.

It remains to establish the equivalence of (b), (c), and (c'). Clearly (c') is
a consequence of (c), applied to/= q>K. Also (c') implies that v*(A) = 0 for
every relatively compact ^-negligible set A; since every /^-negligible set is the
union of relatively compact /^-negligible sets, this proves that (c') implies (b).
To prove that (b) implies (c) we shall argue by contradiction. Suppose there-
fore that there exists a function/0 ^ 0 which is /Mntegrable and v-integrable,
and a real number a > 0 such that, for each integer n > 0, there exists a

function gn for which 0 :g gn ^/0, j gnd\i^2~n and     gn dv > a. By virtue

of (13.5.5), we may replace gn by inf(/0,^) for a suitably chosen function
g'nsJ, without disturbing the above properties, hence (13.9.13) we may
assume that gn is /x-integrable and v-integrable. Now let

h = lim sup gn = inf hn,

n-K                   it


Since gn ^/0 for all w, it follows that hn is ju-integrable and v-integrable
for all n (13.8.2), a$d we have


ld/l- ?1

(13.5.8). Hence |AJ/i = 0 (13.8.1), and the hypothesis (b) therefore implies
that h is v-negligible (13.6.3). But we have f h dv = lim  f hn dv^a (13.8.1),

J                 n-*ao ^

which gives the required contradiction. Hence (b) implies (c), and the proof is

(13.15.6)   Let \i and v be two positive measures on X. Then the following condi-
tions are equivalent:

(a)   The negligible sets are the same for p and for v.it is enough to show that (b) implies that v" -1 = 0. Now the set A